3.52.2 \(\int \frac {e^5 (3-12 x)+x-4 x^2+(-x+2 x^2) \log (x-2 x^2)}{-x^3+2 x^4+e^{10} (-9 x+18 x^2)+e^5 (-6 x^2+12 x^3)} \, dx\)

Optimal. Leaf size=21 \[ \frac {\log (-x (-1+2 x))}{-3 e^5-x} \]

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Rubi [B]  time = 0.76, antiderivative size = 232, normalized size of antiderivative = 11.05, number of steps used = 16, number of rules used = 9, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6688, 6742, 44, 77, 148, 2495, 36, 31, 29} \begin {gather*} -\frac {1+12 e^5}{\left (1+6 e^5\right ) \left (x+3 e^5\right )}+\frac {12 e^5}{\left (1+6 e^5\right ) \left (x+3 e^5\right )}+\frac {1}{\left (1+6 e^5\right ) \left (x+3 e^5\right )}+\frac {2 \log (1-2 x)}{1+6 e^5}-\frac {12 e^5 \log (1-2 x)}{\left (1+6 e^5\right )^2}-\frac {2 \log (1-2 x)}{\left (1+6 e^5\right )^2}-\frac {\log ((1-2 x) x)}{x+3 e^5}+\frac {\left (1+12 e^5+72 e^{10}\right ) \log \left (x+3 e^5\right )}{3 e^5 \left (1+6 e^5\right )^2}-\frac {2 \log \left (x+3 e^5\right )}{1+6 e^5}+\frac {2 \log \left (x+3 e^5\right )}{\left (1+6 e^5\right )^2}-\frac {\log \left (x+3 e^5\right )}{3 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5*(3 - 12*x) + x - 4*x^2 + (-x + 2*x^2)*Log[x - 2*x^2])/(-x^3 + 2*x^4 + E^10*(-9*x + 18*x^2) + E^5*(-6*
x^2 + 12*x^3)),x]

[Out]

1/((1 + 6*E^5)*(3*E^5 + x)) + (12*E^5)/((1 + 6*E^5)*(3*E^5 + x)) - (1 + 12*E^5)/((1 + 6*E^5)*(3*E^5 + x)) - (2
*Log[1 - 2*x])/(1 + 6*E^5)^2 - (12*E^5*Log[1 - 2*x])/(1 + 6*E^5)^2 + (2*Log[1 - 2*x])/(1 + 6*E^5) - Log[(1 - 2
*x)*x]/(3*E^5 + x) - Log[3*E^5 + x]/(3*E^5) + (2*Log[3*E^5 + x])/(1 + 6*E^5)^2 - (2*Log[3*E^5 + x])/(1 + 6*E^5
) + ((1 + 12*E^5 + 72*E^10)*Log[3*E^5 + x])/(3*E^5*(1 + 6*E^5)^2)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 148

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g
, h, m}, x] && (IntegersQ[m, n, p] || (IGtQ[n, 0] && IGtQ[p, 0]))

Rule 2495

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(h*(m + 1)), x] + (-Dist[(b*p*r)/(
h*(m + 1)), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[(d*q*r)/(h*(m + 1)), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^5 (3-12 x)-x+4 x^2-x (-1+2 x) \log \left (x-2 x^2\right )}{(1-2 x) x \left (3 e^5+x\right )^2} \, dx\\ &=\int \left (\frac {1}{\left (3 e^5+x\right )^2 (-1+2 x)}-\frac {4 x}{\left (3 e^5+x\right )^2 (-1+2 x)}-\frac {3 e^5 (-1+4 x)}{x \left (3 e^5+x\right )^2 (-1+2 x)}+\frac {\log ((1-2 x) x)}{\left (3 e^5+x\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {x}{\left (3 e^5+x\right )^2 (-1+2 x)} \, dx\right )-\left (3 e^5\right ) \int \frac {-1+4 x}{x \left (3 e^5+x\right )^2 (-1+2 x)} \, dx+\int \frac {1}{\left (3 e^5+x\right )^2 (-1+2 x)} \, dx+\int \frac {\log ((1-2 x) x)}{\left (3 e^5+x\right )^2} \, dx\\ &=-\frac {\log ((1-2 x) x)}{3 e^5+x}-2 \int \frac {1}{(1-2 x) \left (3 e^5+x\right )} \, dx-4 \int \left (\frac {3 e^5}{\left (1+6 e^5\right ) \left (3 e^5+x\right )^2}-\frac {1}{\left (1+6 e^5\right )^2 \left (3 e^5+x\right )}+\frac {2}{\left (1+6 e^5\right )^2 (-1+2 x)}\right ) \, dx-\left (3 e^5\right ) \int \left (\frac {1}{9 e^{10} x}+\frac {-1-12 e^5}{3 e^5 \left (1+6 e^5\right ) \left (3 e^5+x\right )^2}+\frac {-1-12 e^5-72 e^{10}}{9 e^{10} \left (1+6 e^5\right )^2 \left (3 e^5+x\right )}+\frac {8}{\left (1+6 e^5\right )^2 (-1+2 x)}\right ) \, dx+\int \frac {1}{x \left (3 e^5+x\right )} \, dx+\int \left (-\frac {1}{\left (1+6 e^5\right ) \left (3 e^5+x\right )^2}-\frac {2}{\left (1+6 e^5\right )^2 \left (3 e^5+x\right )}+\frac {4}{\left (1+6 e^5\right )^2 (-1+2 x)}\right ) \, dx\\ &=\frac {1}{\left (1+6 e^5\right ) \left (3 e^5+x\right )}+\frac {12 e^5}{\left (1+6 e^5\right ) \left (3 e^5+x\right )}-\frac {1+12 e^5}{\left (1+6 e^5\right ) \left (3 e^5+x\right )}-\frac {2 \log (1-2 x)}{\left (1+6 e^5\right )^2}-\frac {12 e^5 \log (1-2 x)}{\left (1+6 e^5\right )^2}-\frac {\log (x)}{3 e^5}-\frac {\log ((1-2 x) x)}{3 e^5+x}+\frac {2 \log \left (3 e^5+x\right )}{\left (1+6 e^5\right )^2}+\frac {\left (1+12 e^5+72 e^{10}\right ) \log \left (3 e^5+x\right )}{3 e^5 \left (1+6 e^5\right )^2}+\frac {\int \frac {1}{x} \, dx}{3 e^5}-\frac {\int \frac {1}{3 e^5+x} \, dx}{3 e^5}-\frac {2 \int \frac {1}{3 e^5+x} \, dx}{1+6 e^5}-\frac {4 \int \frac {1}{1-2 x} \, dx}{1+6 e^5}\\ &=\frac {1}{\left (1+6 e^5\right ) \left (3 e^5+x\right )}+\frac {12 e^5}{\left (1+6 e^5\right ) \left (3 e^5+x\right )}-\frac {1+12 e^5}{\left (1+6 e^5\right ) \left (3 e^5+x\right )}-\frac {2 \log (1-2 x)}{\left (1+6 e^5\right )^2}-\frac {12 e^5 \log (1-2 x)}{\left (1+6 e^5\right )^2}+\frac {2 \log (1-2 x)}{1+6 e^5}-\frac {\log ((1-2 x) x)}{3 e^5+x}-\frac {\log \left (3 e^5+x\right )}{3 e^5}+\frac {2 \log \left (3 e^5+x\right )}{\left (1+6 e^5\right )^2}-\frac {2 \log \left (3 e^5+x\right )}{1+6 e^5}+\frac {\left (1+12 e^5+72 e^{10}\right ) \log \left (3 e^5+x\right )}{3 e^5 \left (1+6 e^5\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 19, normalized size = 0.90 \begin {gather*} -\frac {\log ((1-2 x) x)}{3 e^5+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(3 - 12*x) + x - 4*x^2 + (-x + 2*x^2)*Log[x - 2*x^2])/(-x^3 + 2*x^4 + E^10*(-9*x + 18*x^2) + E^
5*(-6*x^2 + 12*x^3)),x]

[Out]

-(Log[(1 - 2*x)*x]/(3*E^5 + x))

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fricas [A]  time = 1.16, size = 18, normalized size = 0.86 \begin {gather*} -\frac {\log \left (-2 \, x^{2} + x\right )}{x + 3 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-x)*log(-2*x^2+x)+(-12*x+3)*exp(5)-4*x^2+x)/((18*x^2-9*x)*exp(5)^2+(12*x^3-6*x^2)*exp(5)+2*x^
4-x^3),x, algorithm="fricas")

[Out]

-log(-2*x^2 + x)/(x + 3*e^5)

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giac [A]  time = 0.16, size = 18, normalized size = 0.86 \begin {gather*} -\frac {\log \left (-2 \, x^{2} + x\right )}{x + 3 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-x)*log(-2*x^2+x)+(-12*x+3)*exp(5)-4*x^2+x)/((18*x^2-9*x)*exp(5)^2+(12*x^3-6*x^2)*exp(5)+2*x^
4-x^3),x, algorithm="giac")

[Out]

-log(-2*x^2 + x)/(x + 3*e^5)

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maple [A]  time = 0.24, size = 19, normalized size = 0.90

method	result	size
norman	\(-\frac {\ln \left (-2 x^{2}+x \right )}{x +3 \,{\mathrm e}^{5}}\)	\(19\)
risch	\(-\frac {\ln \left (-2 x^{2}+x \right )}{x +3 \,{\mathrm e}^{5}}\)	\(19\)
default	\(\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (9 \,{\mathrm e}^{10}+6 \textit {\_Z} \,{\mathrm e}^{5}+\textit {\_Z}^{2}\right )}{\sum }\left (\frac {\underline {\hspace {1.25 ex}}\alpha }{\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+\frac {3 \,{\mathrm e}^{5}}{\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}\right ) \left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (-2 x^{2}+x \right )-\dilog \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )-\dilog \left (\frac {2 x -1}{2 \underline {\hspace {1.25 ex}}\alpha -1}\right )-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {2 x -1}{2 \underline {\hspace {1.25 ex}}\alpha -1}\right )\right )\right )}{18}+\frac {4 \ln \left (x +3 \,{\mathrm e}^{5}\right )}{6 \,{\mathrm e}^{5}+1}+\frac {\ln \left (x +3 \,{\mathrm e}^{5}\right )}{3 \,{\mathrm e}^{5} \left (6 \,{\mathrm e}^{5}+1\right )}-\frac {2 \ln \left (2 x -1\right )}{6 \,{\mathrm e}^{5}+1}-\frac {\ln \relax (x )}{3 \,{\mathrm e}^{5}}\)	\(189\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-x)*ln(-2*x^2+x)+(-12*x+3)*exp(5)-4*x^2+x)/((18*x^2-9*x)*exp(5)^2+(12*x^3-6*x^2)*exp(5)+2*x^4-x^3),
x,method=_RETURNVERBOSE)

[Out]

-ln(-2*x^2+x)/(x+3*exp(5))

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maxima [B]  time = 0.40, size = 303, normalized size = 14.43 \begin {gather*} -\frac {1}{3} \, {\left (e^{\left (-10\right )} \log \relax (x) - \frac {{\left (12 \, e^{5} + 1\right )} \log \left (x + 3 \, e^{5}\right )}{36 \, e^{20} + 12 \, e^{15} + e^{10}} - \frac {36 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {3}{x {\left (6 \, e^{10} + e^{5}\right )} + 18 \, e^{15} + 3 \, e^{10}}\right )} e^{5} - 12 \, {\left (\frac {2 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} - \frac {2 \, \log \left (x + 3 \, e^{5}\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {1}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}}\right )} e^{5} + \frac {1}{3} \, e^{\left (-5\right )} \log \relax (x) - \frac {{\left (12 \, e^{5} + 1\right )} \log \left (x + 3 \, e^{5}\right )}{3 \, {\left (6 \, e^{10} + e^{5}\right )}} - \frac {{\left (6 \, e^{5} + 1\right )} \log \relax (x) - {\left (2 \, x - 1\right )} \log \left (-2 \, x + 1\right )}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} + \frac {12 \, e^{5}}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} - \frac {2 \, \log \left (2 \, x - 1\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {2 \, \log \left (x + 3 \, e^{5}\right )}{36 \, e^{10} + 12 \, e^{5} + 1} + \frac {1}{x {\left (6 \, e^{5} + 1\right )} + 18 \, e^{10} + 3 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-x)*log(-2*x^2+x)+(-12*x+3)*exp(5)-4*x^2+x)/((18*x^2-9*x)*exp(5)^2+(12*x^3-6*x^2)*exp(5)+2*x^
4-x^3),x, algorithm="maxima")

[Out]

-1/3*(e^(-10)*log(x) - (12*e^5 + 1)*log(x + 3*e^5)/(36*e^20 + 12*e^15 + e^10) - 36*log(2*x - 1)/(36*e^10 + 12*
e^5 + 1) + 3/(x*(6*e^10 + e^5) + 18*e^15 + 3*e^10))*e^5 - 12*(2*log(2*x - 1)/(36*e^10 + 12*e^5 + 1) - 2*log(x
+ 3*e^5)/(36*e^10 + 12*e^5 + 1) + 1/(x*(6*e^5 + 1) + 18*e^10 + 3*e^5))*e^5 + 1/3*e^(-5)*log(x) - 1/3*(12*e^5 +
1)*log(x + 3*e^5)/(6*e^10 + e^5) - ((6*e^5 + 1)*log(x) - (2*x - 1)*log(-2*x + 1))/(x*(6*e^5 + 1) + 18*e^10 +
3*e^5) + 12*e^5/(x*(6*e^5 + 1) + 18*e^10 + 3*e^5) - 2*log(2*x - 1)/(36*e^10 + 12*e^5 + 1) + 2*log(x + 3*e^5)/(
36*e^10 + 12*e^5 + 1) + 1/(x*(6*e^5 + 1) + 18*e^10 + 3*e^5)

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mupad [B]  time = 3.62, size = 18, normalized size = 0.86 \begin {gather*} -\frac {\ln \left (x-2\,x^2\right )}{x+3\,{\mathrm {e}}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2 - x + log(x - 2*x^2)*(x - 2*x^2) + exp(5)*(12*x - 3))/(exp(10)*(9*x - 18*x^2) + exp(5)*(6*x^2 - 12*
x^3) + x^3 - 2*x^4),x)

[Out]

-log(x - 2*x^2)/(x + 3*exp(5))

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sympy [A]  time = 0.19, size = 15, normalized size = 0.71 \begin {gather*} - \frac {\log {\left (- 2 x^{2} + x \right )}}{x + 3 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2-x)*ln(-2*x**2+x)+(-12*x+3)*exp(5)-4*x**2+x)/((18*x**2-9*x)*exp(5)**2+(12*x**3-6*x**2)*exp(5
)+2*x**4-x**3),x)

[Out]

-log(-2*x**2 + x)/(x + 3*exp(5))

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